2017 UNM-PNM Statewide High School Mathematics Contest II Problems/Problem 2

Problem

Suppose $A, R, S$ , and $T$ all denote distinct digits from $1$ to $9$ . If $\sqrt{STARS} = SAT$ , what are $A, R, S$ , and $T$ ?

Solution

$A = 3, T = 9, S = 1$ and $R = 2$

Another way of stating the question is that $SAT$ $\cdot$ $SAT$ = $STARS$ . This means that the ending digit of $T$ $\cdot$ $T$ or $S$ has to be the ending digit of a square number, or { $0, 1, 4, 5, 6, 9$ }. Keep in mind that $STARS$ is a 5-digit number. We know that $S$ cannot be 0 because $SAT$ would become a 2-digit number. It cannot be 4 or over because $STARS$ would have 6 digits instead of 5. This only leaves $S$ = 1. Because the last digit of $STARS$ is $1$ , we can conclude that $T$ must be $9$ or $1$ . However, since all the digits are unique, $T$ = $9$ . Now the expression becomes:

$1A9$ $\cdot$ $1A9$ = $19AR1$

Because the first two digits of $STARS$ are $1$ and $9$ , $A$ cannot take on too many values because if it becomes $5$ , $S$ will become $2$ . It cannot be $4$ either because even then, $STARS$ would be too small to satisfy (If $SAT=140$ , it would still have too big a square for $STARS$ to be able to satisfy, $A$ being $4$ ). Now, if $A$ is tested from the set { $0,1,2,3$ }:

$A=0$ :

$109$ $\cdot$ $109$ = $11881$ (Does not work)

$A=1$ :

$119$ $\cdot$ $119$ = $14161$ (Does not work)

$A=2$ :

$129$ $\cdot$ $129$ = $16641$ (Does not work)

$A=3$ :

$139$ $\cdot$ $139$ = $19321$

This works as it fits the template: $\boxed{\textbf{(S=1, T=9, A=3, and R=2)}}$

2017 UNM-PNM Contest II (Problems • Answer Key • Resources)
Preceded by Problem 1	Followed by Problem 3
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10
All UNM-PNM Problems and Solutions

Page

Toolbox

Search

2017 UNM-PNM Statewide High School Mathematics Contest II Problems/Problem 2

Problem

Solution

See also